Write the equation of the normal to the curve `y=x+sinxcosx` at `x=pi/2` .

Options

A. `x=2`
B. `x=pi`
C. `x+pi=0`
D. `2x=pi`

Answer

D

Explanation

When `x=(pi)/(2)`, we have
`y=(pi)/(2) + "sin" (pi)/(2)"cos"(pi)/(2)=(pi)/(2)`
So, the coordinates of the point are `(pi//2,pi//2)`
Now,
`y=x+sin x cos x `
`rArr (dy)/(dx)=1+cos^(2)x-sin^(2)x rArr ((dy)/(dx))_(x=(n)/(2))=1+0-1=0`
This means that the tangent at `(pi//2,pi//2)` is parallel to x-axis. So, the normal is parallel to y-axis and hence its equation is `x=(pi)/(2)`.